This is related to Hatcher's Algebraic Topology Example 0.15. Denote $I=[0,1]$.
Given $I\times I$, I can induce retract $I\times I\to I\times\{0\}\cup \partial I\times I$. The retract can be obtained by shining light at $(1/2,1)$ position to the edge $I\times\{0\}\cup\partial I\times I$ and the ray defines the whole ray being collapsed to the target point on $I\times\{0\}\cup\partial I\times I$. This is clearly continuous retract map.
Then given $f:B\to A$, one can have $B\times I\times I\to B\times I\times\{0\}\cup B\times\partial I\times I$.
The book then says this retraction induces $M_f\times I\to M_f\times\{0\}\cup(A\cup B)\times I$. It seems that the book is gluing $B\times 1\times 0$ to $A\times 1\times 0$ part via $f$. Here I will make disjoint union explicit. However, $B\times I\times\{0\}\sqcup B\times\partial I\times I\sqcup A\times 1 \times 0/\sim=(M_f\times\{0\}\cup A\times 1\times I)\sqcup B\times 0\times I$. Here $\cup$ is taken over the image set.
$\textbf{Q:}$ Does the book mean $A\times 1\times I\sqcup B\times 0\times I=(A\sqcup B)\times I$? I have to note that $M_f\times 0$ contains $A\times 1\times 0\subset A\times 1\times I$ part. Or am I wrong on this part?
Yes. The mapping cylinder of $f :B \to A$ is the quotient space $B \times I \sqcup A / \sim$ where $\sim$ is generated by $(a,0) \sim f(a)$. There are canonical embeddings $i_A : A \to M_f, i_A(a) = [a]$ and $i_B : B \to M_f, i_b(b) = [b,1]$. In that sense $A$ is identified with the base $i_A(A)$ of $M_f$ and $B$ is identified with the top $i_B(B)$ of $M_f$.
If $p : B \times I \sqcup A \to M_f$ denotes the quotient map, then $A \sqcup B$ stands for $p(B \times \{ 1 \} \sqcup A)$.